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I have a 24Vac, 50Hz source and RGB LED to be driven from it. As the LED reverse max is like 5V, I have to prevent the reverse half cycle of voltage from being applied across the LED. There are two ways we can connect the reverse protection diodes, either in series or parallel with the LED.

schematic

simulate this circuit – Schematic created using CircuitLab

  1. Now for my case I would prefer to have it in series to reduce cost, but I think this is not such a good way as in the reverse cycle, both D1 and D2 will be off and we cant be sure of how the reverse voltage will get divided across the two. Understood the leakage current would be very small.

  2. If I use a Schottky for the D2 or a silicon diode for D2, I thought the reverse voltage across D1 wouldn't be that much different. As I was not sure, I did a simple Ltspice sim and the result was quite unexpected (probably only for me) enter image description here As far as the sim is concerned, with the Schottky, the LED will be subjected to approx -32V (V_D2) whereas with the Silicon diode it would not (V_D4).

How would you explain this?

Cheers, Kay

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  • \$\begingroup\$ I have a feeling that Q2 behavior is due to the SPICE modeling of Schottky diode Vs Silicon diode. When I add a 3k3 resistor from V_D2 and VD_4 nodes to GND, I get pretty much expected results. \$\endgroup\$
    – Kaush
    Mar 11 at 6:09
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    \$\begingroup\$ In the case with the Schottky, if you use it in practice, do you intend on including both 3k3 resistors? Do you understand why the 3k3 addition did what it did in simulation? \$\endgroup\$ Mar 11 at 6:30
  • \$\begingroup\$ FYI, most Schottky diodes are Silicon. \$\endgroup\$ Mar 11 at 15:13

2 Answers 2

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The result is not unsurprising and is explained with leakage current.

In reverse bias, Schottky diodes pass significantly more current that standard silicon diodes.

By adding the 3k3 resistor as load, even 1 mA of reverse leakage will only result into 3.3V in reverse over the resistor and LED.

But at higher temperature and higher voltage, that specific Schottky could leak up to 10mA.

So it might not be that the model for the Schottky is incorrect.

It can also be a non-realistic LED model so you may get non-realistic results for the LED reverse bias. You would need to measure the current in simulation and see how much it is, but on the other hand, you likely cannot know detailed model for your specific LED, but simply a specification of max reverse voltage Vr e.g. 5V and max reverse current measured at Vr e.g. 100uA at 5V.

If you really want to protect the LED from reverse bias, put another diode in anti-parallel with it. It could be another LED if the LED forward voltage is less than allowed LED reverse voltage.

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    \$\begingroup\$ I don't think the OP actually intended on adding that 2nd 3k3 resistor. I think the OP just poked it into simulation after seeing earlier bad news, saw that adding it made things look better and more as expected, and so then decided that the original circuit without the added 3k3 was just fine, Schottky leakage ignored and without the added 3k3. As you point out, even with the 3k3 added there may be problems. But I don't think the OP intended on keeping it, anyway. Too expensive from what was written there. \$\endgroup\$ Mar 11 at 7:04
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The Schottky diode will have significant reverse leakage current, and way more at high temperature. Here is the typical behavior of the Schottky diode you used:

enter image description here

Keep in mind that at the one data point specified (-60V and 25°C) the guaranteed leakage can be 4x higher than typical. So at -32V and (say) 75°C maximum we might see more than 1mA reverse current.

There is no specification for the long term allowable reverse current on most LED datasheets. From some heavily Muntzed circuits it appears that some LED colors or technologies are more tolerant than others but I don't think we want to depend on that.

The LED model does not model reverse breakdown so you won't be able to simulate the reverse current without changes.

There's also the matter of capacitance. The two devices are roughly similar in capacitance (the diode actually has more) so suddenly turning on AC near the peak will put a nice spike across the LED and a significant current spike if/when it breaks down.

So there are several 100% safe solutions. Diode in parallel (doesn't matter which type, use whichever is cheaper or already on your BOM), which requires the resistor power rating be doubled (wasting some power).

Secondly @justme's suggested solution of one series and one parallel diode and the lower power resistor is 100% safe and compact.

A third solution would be to use an AC LED with two back-to-back dies inside the package. Then increase the resistor value since you're getting light on both half cycles. That's a somewhat oddball LED and may be unattractive for that reason.

Finally you could just use a full bridge rectifier (a single component, if you like) and the higher value resistor and a normal LED.

The third and fourth solutions also produce a less flickery LED appearance since 50Hz half-wave has noticeable flicker.

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